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A292443 a(n) = (5/32)*A000045(6*n)^2. 2
0, 10, 3240, 1043290, 335936160, 108170400250, 34830532944360, 11215323437683690, 3611299316401203840, 1162827164557749952810, 374426735688279083601000, 120564246064461307169569210, 38821312806020852629517684640, 12500342159292650085397524884890 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,2
COMMENTS
Every term is a triangular number. [Problem B-967 in Euler and Sadek, 2003; solution in Euler and Sadek, 2004]
LINKS
Russ Euler and Jawad Sadek, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 41, No. 5 (2003), pp. 466-471.
Russ Euler and Jawad Sadek, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 42, No. 3 (2004), pp. 277-282.
FORMULA
From Colin Barker, Sep 16 2017: (Start)
G.f.: 10*x*(1 + x) / ((1 - x)*(1 - 322*x + x^2)).
a(n) = ((161+72*sqrt(5))^(-n)*(-1+(161+72*sqrt(5))^n)^2) / 32.
a(n) = 323*a(n-1) - 323*a(n-2) + a(n-3) for n > 2.
(End)
a(n) = A000217(A132584(n)). - Amiram Eldar, Jan 11 2022
MATHEMATICA
Table[(5/32) Fibonacci[6 n]^2, {n, 0, 13}] (* Michael De Vlieger, Sep 16 2017 *)
PROG
(PARI) a(n) = (5/32)*fibonacci(6*n)^2
(Magma) [5*Fibonacci(6*n)^2/32: n in [0..20]]; // G. C. Greubel, Feb 03 2019
(Sage) [5*fibonacci(6*n)^2/32 for n in (0..20)] # G. C. Greubel, Feb 03 2019
(GAP) List([0..20], n-> 5*Fibonacci(6*n)^2/32) # G. C. Greubel, Feb 03 2019
CROSSREFS
Subsequence of A000217.
Sequence in context: A225764 A243008 A133198 * A001329 A007101 A007103
KEYWORD
nonn,easy
AUTHOR
Felix Fröhlich, Sep 16 2017
STATUS
approved

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Last modified March 28 07:33 EDT 2024. Contains 371235 sequences. (Running on oeis4.)